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CHAPTER 2
2.1
Based upon Table 2.1, a resistivity of 2.83 -cm < 1 m-cm, and aluminum is a conductor.
2.2
Based upon Table 2.1, a resistivity of 1015 -cm > 105 -cm, and silicon dioxide is an insulator.
2.3
Imax 107 A
cm2
5m 1m
108cm2
m2
500 mA
2.4
a R L
A 2.82x106 cm
2 2 cm
5x104cm 1x104cm 160
b R L
A 2.82x106 cm
2 2 cm
5x104cm 0.5x104cm 319
2.5
a R L
A 1.66x106 cm
2 2 cm
5x104cm 1x104cm 93.9
b R L
A 1.66x106 cm
2 2 cm
5x104cm 0.5x104cm 188
2.6
Tx
EBTn Gi 5
3
1062.8exp
For silicon, B = 1.08 x 1031
and EG = 1.12 eV:
ni = 5.07 x10-19
/cm3 6.73 x10
9/cm
3 8.36 x 10
13/cm
3.
For germanium, B = 2.31 x 1030
and EG = 0.66 eV:
ni = 2.63 x10-4
/cm3 2.27 x10
13/cm
3 8.04 x 10
15/cm
3.
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2.7 Define an M-File:
function f=temp(T)
ni=1E14;
f=ni^2-1.08e31*T^3*exp(-1.12/(8.62e-5*T));
ni = 1013
/cm3 for T = 436 K ni = 10
15/cm3 for T = 602 K
2.8
ni BT3 exp
EG
8.62x105T
with B = 1.27x10
29 K3cm6
T = 300 K and EG = 1.42 eV: ni = 2.21 x106/cm
3
T = 100 K: ni = 6.03 x 10-19/cm
3 T = 450 K: ni = 3.82 x10
10/cm
3
2.9
ni2 BT 3 exp
EG
kT
B = 1.08x10
31
1010 2
1.08x1031T 3 exp 1.12
8.62x105T
Using a spreadsheet, solver, or MATLAB yields T = 305.22K
Define an M-File:
function f=temp(T)
f=1e20-1.08e31*T^3*exp(-1.12/(8.62e-5*T));
Then: fzero('temp',300) | ans = 305.226 K
2.10
vn nE 700cm2
V s
2500
V
cm
1.75x10
6 cm
s
v p pE 250cm2
V s
2500
V
cm
6.25x10
5 cm
s
jn qnvn 1.60x1019C 1017
1
cm3
1.75x10
6 cm
s
2.80x10
4 A
cm2
jp qnv p 1.60x1019C 103
1
cm3
6.25x10
5 cm
s
1.00x10
10 A
cm2
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2.11
jn qnvn 1.60x1019C 1018
1
cm3
10
7 cm
s
1.60x10
6 A
cm21.60
MA
cm2
jp qnvp 1.60x1019C 102
1
cm3
10
7 cm
s
1.60x10
10 A
cm2
2.12
v j
Q2000A / cm2
0.01C / cm2 2x105
cm
s
2.13
22
67
34104
sec104.0
cm
MA
cm
Ax
cm
cm
CQvj
2.14
vn nE 1000cm2
V s
2000
V
cm
2.00x10
6 cm
s
v p pE 400cm2
V s
2000
V
cm
8.00x10
5 cm
s
jn qnvn 1.60x1019C 103
1
cm3
2.00x10
6 cm
s
3.20x10
10 A
cm2
jp qnvp 1.60x1019C 1017
1
cm3
8.00x10
5 cm
s
1.28x10
4 A
cm2
2.15
a E 5V
10x104cm 5000
V
cm b V 105
V
cm
10x10
4cm 100 V
2.16
jp qpv p 1.60x1019C
1019
cm3
10
7 cm
s
1.60x10
7 A
cm2
ip jp A 1.60x107 A
cm2
1x10
4cm 25x104cm 4.00 A
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2.17
For intrinsic silicon, q nni pni qni n p
105 cm 1
for an insulator
ni
q n p
105 cm 1
1.602x1019C 2000 750 cm2
v sec
2.270x1010
cm3
n i2
5.152x1020
cm6 BT 3 exp
EGkT
with
B 1.08x1031 K3cm6, k = 8.62x10-5eV/K and E G 1.12eV
Using MATLAB as in Problem 2.5 yields T ≤ 316.6 K.
2.18
For intrinsic silicon, q nni pni qni n p
1000 cm 1
for a conductor
ni
q n p
1000 cm 1
1.602x1019C 100 50 cm2
v sec
4.16x1019
cm3
n i2
1.73x1039
cm6 BT 3 exp
EG
kT
with
B 1.08x1031 K3cm6, k = 8.62x10-5eV/K and E G 1.12eV
This is a transcendental equation and must be solved numerically by iteration. Using the HP
solver routine or a spread sheet yields T ≥ 2701 K. Note that this temperature is far above the
melting temperature of silicon.
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2.19
Si Si Si
SiB
Si Si Si
P
Dono r electro n
fills accepto r
vacancy
No free electrons or holes (except those corresponding to ni).
2.20 Since Ge is from column IV, acceptors come from column III and donors come from column
V. (a) Acceptors: B, Al, Ga, In, Tl (b) Donors: N, P, As, Sb, Bi
2.21 (a) Gallium is from column 3 and silicon is from column 4. Thus silicon has an extra electron
and will act as a donor impurity.
(b) Arsenic is from column 5 and silicon is from column 4. Thus silicon is deficient in one
electron and will act as an acceptor impurity.
2.22 (a) Germanium is from column IV and indium is from column III. Thus germanium has one
extra electron and will act as a donor impurity.
(b) Germanium is from column IV and phosphorus is from column V. Thus germanium has
one less electron and will act as an acceptor impurity.
2.23
field. electric small a ,20002.0100002 cm
Vcm
cm
Aj
jE
2.24
jndrift
qnnE qnvn 1.602x1019 1017
C
cm3
10
7 cm
s
160
kA
cm2
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2.25
N 1016 atoms
cm3
0.5m 5m 0.5m
104cm
m
3
12,500 atoms
2.26 (a) Since arsenic is a donor, ND = 3 x 1017/cm3. Assume NA = 0, since it is not specified. The
material is n-type.
(b) At room temperature, n i 1010 / cm3 and ND N A 3x10
17 / cm3 >> 2n i
So n 3x1017 /cm3 and p ni
2
n
1020 /cm6
3x1017 /cm3 333/cm3
(c) At 250K, ni2 1.08x1031 250
3
exp 1.12
8.62x105 250
4.53x1015 /cm6
ni 6.73x107 /cm3 N D N A 2ni, so n = 3x10
17 / cm3 and n =4.53x1015
3x1017 0.0151/ cm3
2.27 (a) Since boron is an acceptor, NA = 6 x 1018/cm3. Assume ND = 0, since it is not specified.
The material is p-type.
(b) At room temperature, ni 1010 /cm3 and N A ND 6x10
18 / cm3 >> 2n i
So p 6x1018 /cm3 and n ni
2
p
1020 /cm6
6x1018 /cm316.7/cm3
(c) At 200K, ni2 1.08x1031 200
3
exp 1.12
8.62x105 200
5.28x109 /cm6
ni 7.27x104 /cm3 N A ND 2ni, so p 6x10
18 /cm3 and n 5.28x109
6x1018 8.80x1010 /cm3
2.28 (a) Arsenic is a donor, and boron is an acceptor. ND = 2 x 1018/cm3, and NA = 8 x 1018/cm3.
Since NA > ND, the material is p-type.
3
318
6202318
i
318310
i
716106
10 and 106 So
2n >> /106 and /10n re, temperaturoomAt (b)
/cm./cmx
/cm
p
nn/cmxp
cmxNNcm
i
DA
2.29 (a) Phosphorus is a donor, and boron is an acceptor. ND = 2 x 1017/cm3, and NA = 5 x 1017/cm3.
Since NA > ND, the material is p-type.
3
317
6202317
i
317310
333103
10 and 103 So
2n >> /103 and 10 re, temperaturoomAt (b)
/cm/cmx
/cm
p
nn/cmxp
cmxNN/cmn
i
DAi
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2.30
N A ND: N A ND 5x1016 1016 4x1016 /cm3 2ni 2x10
11 /cm3
p N A ND 4x1016 /cm3 | n
ni2
p
1022
4x1016 2.50x105 /cm3
The material is p - type.
2.31
N A N D: The material is p - type. N A N D 1015 1014 9x1014 /cm3
If we assume N A N D 2ni 1014 / cm3 :
p N A N D 9x1014 /cm3 | n
ni2
p
251026
9x1014 2.78x1012 /cm3
If we use Eq. 2.12 : p 9x1014 9x1014
2
4 5x1013 2
2 9.03x1014
and n 2.77x1012 /cm3. The answers are essentially the same.
2.32
N D N A: The material is n - type. N D N A 3x1017 2x1017 1x1017 /cm3
2ni 2x1017 /cm3; Need to use Eq. (2.11)
n 1017 1017
2
4 1017 2
21.62x1017 /cm3
p ni
2
n
1034
1.62x1017 6.18x1016 /cm3
2.33
ND = 5 x 1016/cm3. Assume NA = 0, since it is not specified.
N D N A : The material is n - type. | N D N A 5x1016 / cm3 2ni 2x10
10 / cm3
n = 5x1016 / cm3 | p =ni
2
n
1020
5x1016 2x103 / cm3
N D N A 5x1016 / cm3 | Using Fig. 2.8, n 960
cm2
V s and p 280
cm2
V s
1
qnn
1
1.602x1019C 960cm2
V s
5x1016
cm3
0.130 cm
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2.34
NA = 2.5x1018/cm3. Assume ND = 0, since it is not specified.
N A N D : The material is p - type. | N A N D 2.5x1018 / cm3 2ni 2x10
10 / cm3
p = 2.5x1018 / cm3 | n =ni
2
p
1020
2.5x1018 40/ cm3
N D N A 2.5x1018 / cm3 | Using Fig. 2.8, n 170
cm2
V s and p 80
cm2
V s
1
qp p
1
1.602x1019C 80cm2
V s
2.5x1018
cm3
31.2 m cm
2.35
Indium is from column 3 and is an acceptor. NA = 8 x 1019/cm3. Assume ND = 0, since it is not
specified.
N A N D : material is p - type | N A N D 8x1019 /cm3 2ni 2x10
10 /cm3
p 8x1019 /cm3 | n=ni
2
p
1020
8x10191.25/cm3
N D N A 7x1019 / cm3 | Using Fig. 2.8, n 100
cm2
V s and p 50
cm2
V s
1
q p p
1
1.602x1019C 50cm2
V s
8x1019
cm3
1.56 m cm
2.36
Phosphorus is a donor : ND 4.5x1016 / cm3 | Boron is an acceptor : N A 5.5x10
16 / cm3
N A N D : The material is p - type. | N A N D 1016 / cm3 2ni 2x10
10 / cm3
p 1016 /cm3 | n ni
2
p
1020
1016104 /cm3
N D N A 1017 / cm3 | Using Fig. 2.8, n 800
cm2
V s and p 220
cm2
V s
1
qnn
1
1.602x1019C 220cm2
V s
1016
cm3
2.84 cm
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2.37
1
qp p | p p
1
1.602x1019C 0.054 cm
1.16x1020
V cm s
An iterative solution is required. Using Fig. 2.8:
NA p p p
1018 100 1.0 x 1020
1.1 x1018 100 1.1 x 1020
1.2 x 1017 95 1.14 x 1020
1.3 x 1019 90 1.17 x 1020
2.38
1
qnn | nn
1
1.602x1019C 0.054 cm
1.16x1020
V cm s
An iterative solution is required. Using the equations in Fig. 2.8:
ND n n n
1016 1250 1.25 x 1019
1018 264 2.64 x 1020
1017 802 8.02 x 1019
1.2x1017 604 1.21 x 1020
1.85 x 1019 626 1.16 x 1020
2.39
1
qp p | p p
1
1.602x1019C 0.054 cm
1.16x1020
V cm s
An iterative solution is required. Using the equations in Fig. 2.8:
NA p p p
1018 96.7 9.67 x 1020
1.1 x1018 93.7 1.03 x 1020
1.2 x 1017 91.0 1.09 x 1020
1.3 x 1019 88.7 1.15 x 1020
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2.40
1
qp p | p p
1
1.602x1019C 0.5 cm
1.25x1019
V cm s
An iterative solution is required. Using the equations in Fig. 2.8:
NA p p p
1016 406 4.06 x
1018
2 x 1016 363 7.26 x
1018
3 x 1016 333 9.99 x
1018
4 x 1016 310 1.24 x 1019
2.41 Yes, by adding equal amounts of donor and acceptor impurities the mobilities are reduced, but
the hole and electron concentrations remain unchanged. See Problem 2.44 for example.
However, it is physically impossible to add exactly equal amounts of the two impurities.
2.42
1
qnn | nn nND
1
1.602x1019C 3 cm
2.08x1018
V cm s
An iterative solution is required. Using the equations in Fig. 2.8:
ND n nn
1015 1350 1.35 x 1018
1.5 x 1015 1340 2.01 x 1018
1.6 x 1015 1340 2.14 x 1018
1.55 x 1015 1340 2.08 x 1018
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2.43 (a)
1
qnn | nn nND
1
1.602x1019C 0.001 cm
6.24x1021
V cm s
An iterative solution is required. Using the equations in Fig. 2.8:
ND n nn
1019 116 1.16 x 1021
7 x 1019 96.1 6.73 x 1021
6.5 x 1019 96.4 6.27 x 1021
(b)
1
qp p | p p pNA
1
1.602x1019C 0.001 cm
6.24x1021
V cm s
An iterative solution is required using the equations in Fig. 2.8:
NA p p p
1 x1020 49.6 4.96 x 1021
1.2 x1020 49.4 5.93 x 1021
1.25 x1020 49.4 6.17 x 1021
1.26 x 1020 49.4 6.22 x 1021
2.44
Based upon the value of its resistivity, the material is an insulator. However, it is not intrinsic
because it contains impurities. Addition of the impurities has increased the resistivity.
Since N D N A = 0, n = p = ni , and q nni pni qni n p N A N D 10
20 / cm3 which yields p 49.6 and n 95.0 using the
equations from Fig. 2.8.
1
qni n p
1
1.602x1019C 1010cm3 95.0 49.6 cm2
v sec
4.32x106 - cm
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2.45
(a) For the 1 ohm-cm starting material:
1
qp p | p p pN A
1
1.602x1019C 1 cm
6.25x1018
V cm s
An iterative solution is required. Using the equations in Fig. 2.8:
NA p p p
1016 406 4.1 x 1018
1.5 x 1016 383 5.7 x 1018
1.7 x 1016 374 6.4 x 1019
To change the resistivity to 0.25 ohm-cm:
1
qp p | p p pN A
1
1.602x1019C 0.25 cm
2.5x1019
V cm s
NA p p p
6 x 1016 276 1.7 x 1019
8 x 1016 233 2.3 x 1019
1.1 x 1017 225 2.5 x 1019
Additional acceptor concentration = 1.1x1017
- 1.7x1016
= 9.3 x 1016
/cm3
(b) If donors are added:
ND ND + NA n ND - NA nn
2 x 1016 3.7 x 1016 1060 3 x 1015 3.2 x 1018
1 x 1017 1.2 x 1017 757 8.3 x 1016 6.3 x 1019
8 x 1016 9.7 x 1016 811 6.3 x 1016 5.1 x 1019
4.1 x 1016 5.8 x 1016 950 2.4 x 1016 2.3 x 1019
So ND = 4.1 x 1016
/cm3 must be added to change achieve a resistivity of 0.25 ohm-cm. The
silicon is converted to n-type material.
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2.46
Phosphorus is a donor: ND = 1016/cm3 and n = 1250 cm2/V-s from Fig. 2.8.
qnn qnND 1.602x1019C 1250 1016
2.00
cm
Now we add acceptors until = 5.0 (-cm) -1:
qp p | p p p N A ND 4 cm
1
1.602x1019C
2.50x1019
V cm s
NA ND + NA p NA - ND p p
1.00E+17 1.10E+17 2.25E+02 9.00E+16 2.02E+19
2.00E+17 2.10E+17 1.76E+02 1.90E+17 3.34E+19
1.50E+17 1.60E+17 1.95E+02 1.40E+17 2.74E+19
1.40E+17 1.50E+17 2.00E+02 1.30E+17 2.60E+19
1.30E+17 1.40E+17 2.06E+02 1.20E+17 2.47E+19
1.32E+17 1.42E+17 2.05E+02 1.22E+17 2.50E+19
2.47
Boron is an acceptor: NA = 1016/cm3 and p = 405 cm2/V-s from Fig. 2.8.
qp p qpNA 1.602x1019C 405 1016
0.649
cm
Now we add donors until = 5.5 (-cm) -1:
qnn | nn n ND N A 5.5 cm
1
1.602x1019C
3.43x1019
V cm s
ND ND + NA n ND - NA p p
8 x 1016 9 x 1016 832 7 x 1016 5.8 x 1019
6 x 1016 7 x 1016 901 5 x 1016 4.5 x 1019
4.5 x 1016 5.5 x 1016 964 3.5 x 1016 3.4 x 1019
2.48
VT kT
q
1.38x1023T
1.602x1019 8.62x105T
T (K) 50 75 100 150 200 250 300 350 400
VT (mV) 4.31 6.46 8.61 12.9 17.2 21.5 25.8 30.1 34.5
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2.49
j qDn dn
dx
qVTn
dn
dx
j 1.602x1019C 0.025V 350cm2
V s
1018 0
0 0.5x104
1
cm4 28.0
kA
cm2
2.50
j qDpdp
dx 1.602x1019C 15
cm2
s
1019 / cm3
2x104cm
exp
x
2x104cm
j 1.20x105 exp 5000x
cm
A
cm2
I 0 j 0 A 1.20x105A
cm2
10m
2 108cm2
m2
12.0 mA
2.51
jp q p pE qDpdp
dx q p p E VT
1
p
dp
dx
0 E VT
1
p
dp
dx
E VT1
N A
dN Adx
0.0251022 exp 104 x
1014 1018 exp 104 x
E 0 0.0251022
1014 1018 250
V
cm
E 5x104cm 0.0251022 exp 5
1014 1018 exp 5 246
V
cm
2.52
At x = 0:
jndrift
qnnE 1.60x1019C 350
cm2
V s
1016
cm3
20
V
cm
11.2
A
cm2
jpdrift
q p pE 1.60x1019C 150
cm2
V s
1.01x1018
cm3
20
V
cm
484
A
cm2
jndiff
qDndn
dx 1.60x1019C 350 0.025
cm2
s
104 1016
2x104cm4
70.0
A
cm2
jpdiff
qDpdp
dx 1.60x1019C 150 0.025
cm2
s
1018 1.01x1018
2x104cm4
30.0
A
cm2
jT 11.2 484 70.0 30.0 535A
cm2
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At x = 1 m assuming linear distributions :
p 1m 1.005x1018 / cm3 , n 1m 5x1015 / cm3
jndrift
qnnE 1.60x1019C 350
cm2
V s
5x1015
cm3
20
V
cm
5.61
A
cm2
jpdrift
q p pE 1.60x1019C 150
cm2
V s
1.005x1018
cm3
20
V
cm
482
A
cm2
jndiff
qDndn
dx 1.60x1019C 350 0.025
cm2
s
104 1016
2x104cm4
70.0
A
cm2
jpdiff
qDpdp
dx 1.60x1019C 150 0.025
cm2
s
1018 1.01x1018
2x104cm 4
30.0
A
cm2
jT 5.61 482 70.0 30.0 528A
cm2
2.53 NA = 2ND
EC
EA
NA
Holes
EV
NA
ND
NA
ND
NA
ND
ED
2.54
hc
E
6.626x1034 J s 3x108 m / s 1.12eV 1.602x1019 J / eV
1.108 m
2.55
p-type silicon
n-type silicon
Si02
Al - Anode Al - Cathode
-
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©R. C. Jaeger & T. N. Blalock 10/18/09 2-16
2.56 An n-type ion implantation step could be used to form the n+ region following step (f) in Fig.
2.17. A mask would be used to cover up the opening over the p-type region and leave the
opening over the n-type silicon. The masking layer for the implantation could just be
photoresist.
n-type silicon
p-type silicon
Si02
Photoresist
Structure after exposure and development of photoresist layer
Mask
n-type silicon
p-type silicon
Structure following ion implantation of n-type impurity
n+
Ion implantation
Side view
Top View
Mask for ion implantation
2.57
(a) N = 81
8
6
1
2
4 1 8 atoms
(b) V l 3 0.543x109 m 3
0.543x107cm 3
1.60x1022cm3
(c) D =8 atoms
1.60x1022cm3 5.00x1022
atoms
cm3
(d ) m = 2.33g
cm3
1.60x10
22cm3 3.73x1022 g
(e) From Table 2.2, silicon has a mass of 28.086 protons.
mp 3.73x1022 g
28.082 8 protons1.66x1024
g
proton
Yes, near the actual proton rest mass.